Pi appears a LOT in trigonometry, but only because of its 'circle-significance'. Does pi ever matter in things not concerned with circles? Is its only claim to fame the fact that its irrational and an important ratio?

I'd rather this be a comment instead, so: $\pi$ turns up in the expression for the so-called "probability integral" (a.k.a. the "error function") among other things. How circles relate to this is a bit of a long-winded explanation though.
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Guess who it is.Aug 26 '10 at 0:11

Also, let's get one thing straight here: circles are eerily important. You will never stop running into circles in mathematics.
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Qiaochu YuanAug 26 '10 at 0:26

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(For example, although the Fourier transform is "concerned with circles" (functions on the circle being the same thing as periodic functions) it penetrates into the deepest parts of modern mathematics. Many appearances of pi are because of a Fourier transform lurking somewhere in the background. You might also want to read this MO thread where I asked a similar question: mathoverflow.net/questions/18180/…)
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Qiaochu YuanAug 26 '10 at 0:52

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Fundamental source of $\pi$ is circle nothing else. It may be difficult to find it but it is always there.
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Pratik DeoghareAug 26 '10 at 12:43

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It is difficult to know if a circle is not lurking somewhere, whenever there is $\pi$, but the values of the Riemann zeta function at the positive even integers have a lot to do with powers of $\pi$: see here for the values.

For instance, you can prove that the probability that two "randomly chosen" positive integers are coprime is $\frac{1}{\zeta(2)} = \frac{6}{\pi^2}$.

@asmeurer: True, but I'm saying it has nothing to do with "angles and the 2D lattice". Nothing I know of, anyway. The proof that $\zeta(2)$ has that value (see the linked article) doesn't use those, and the proof that the probability of two "randomly chosen" prime numbers is $\frac{1}{\zeta(2)}$ is straightforward.
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ShreevatsaRJun 10 '13 at 2:01

$\pi$ appears in Stirling's approximation, which is not obviously related to circles. This means that $\pi$ appears in asymptotics related to binomial coefficients, such as

$$\displaystyle {2n \choose n} \approx \frac{4^n}{\sqrt{\pi n}}.$$

In other words, the probability of flipping exactly $n$ heads and $n$ tails after flipping a coin $2n$ times is about $\frac{1}{\sqrt{\pi n}}$. This asymptotic also suggests that on average you should flip between $n + \sqrt{\pi n}$ and $n - \sqrt{\pi n}$ heads.

This is closely related to J. Mangaldan's comment about the probability integral. Somehow I think it all ties back to the fact that e^{-x^2} is its own Fourier transform.
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Qiaochu YuanAug 26 '10 at 0:31

I guess that comment is worth explaining: the relationship is that the constant in Stirling's approximation can be computed from the central limit theorem. This is explained at terrytao.wordpress.com/2010/01/02/… .
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Qiaochu YuanJan 14 '11 at 0:53

@QiaochuYuan You might be interested in Kunth's "Why Pi?" Lecture. He shows how this is related to cirlces!
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Pedro Tamaroff♦Feb 27 '12 at 5:19

Yes, the ratio $\pi$ of a circle's circumference to its diameter shows up in many, many places where one might not expect it!

One partial explanation (similar in spirit to "circles lurk everywhere") is that the equation for a circle is a $quadratic$ (eg. $x^2+y^2 = r^2$.) After nice linear functions, the next most commonly used functions are quadratic functions and everywhere one runs into a quadratic function, a trig substitution (e.g. $x = r \cos \theta; y=r\sin \theta$) may be useful, turning the quadratic function into something involving $\pi.$ This explains the antiderivative $\int \frac{1}{1+x^2} dx$ involving $\pi$, the sum of reciprocals of squares $\sum^\infty\frac{1}{k^2}$ involving $\pi$ and the area under the Gaussian distribution involving $\pi$. And so on....

@Qiaochu: Most of the proofs I know apply equally well to evaluating $\sum 1/n^d$ (for d even) and even $\sum (-1)^d/n^d$ (for d odd), which also involve $\pi$. So, the fact that the terms are squares doesn't seem particularly significant to the appearance of $\pi$.
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George LowtherJan 14 '11 at 1:41

I'll have a look through the alternative proofs in that link though.
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George LowtherJan 14 '11 at 1:42

(I meant $\sum(-1)^d/(2n+1)^d$ above). I always thought of these sums involving $\pi$ for similar reasons, and not just the $d=2$ case in isolation.
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George LowtherJan 14 '11 at 1:56